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The question is, "Determine whether each of these arguments is valid. If an argument is correct, what rule of inference is being used? If it is not, what logical error occurs?

a) If $n$ is a real number such that $n>1$, then $n^2>1$. Suppose that $n^2>1$. Then $n>1$.

b)If $n$ is a real number with $n>3$, then $n^2>9$. Suppose that $n^2≤9$. Then $n ≤3$.

c) If $n$ is a real number with $n>2$, then $n^2>4$. Suppose that $n≤2$. Then $n^2≤4$.

The only one I am having difficulty with is part c). The answer to part c), is that it is a fallacy of denying the hypothesis. I thought, "perhaps by taking the contrapositive of the conditional statement, it might make things a little more lucid." However, I was wrong on that idea. If someone could explain part c), I'd be so quite grateful. Thank you!

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up vote 4 down vote accepted

You mention you're "okay with" (a) and (b):

(a) Just to double-check, you should have seen that it is not valid, as it asserts the converse of the first statement, $q\rightarrow p$, which is not equivalent to the first statement of the form $p \rightarrow q$.

  • The inverse of $\;p\rightarrow q\;$ is $\;\lnot p \rightarrow \lnot q$, which is equivalent to (is the contrapositive of), the the converse of the implication. Neither the converse of an implication nor its inverse are equivalent to the original implication.

(b) is valid, as it states the equivalent in the form of the contrapositive of the first statement. That is, if we are given $p \rightarrow q$, then it is validly asserting $\lnot q \rightarrow \lnot p$.

c) If $n$ is a real number with $n>2$, then $n^2>4$. Suppose that $n≤2$. Then $n^2≤4$.

Let "$n$ is a real number with $n > 2$" be denoted by $p$.

Let "$n^2>4$" be denoted by $q$.

Then $n \leq 4 \equiv \lnot p$. The negation of "$n> 2$" is " n is not greater than 2, i.e., $n \leq 2$.

And $n^2 \leq 4 \equiv \lnot q$. Can you see why? The negation of "$n^2 > 4$ is "$n^2$ is not greater than 4", i.e., $n^2 \leq 4$.

The first statement in (c) is of the form $p\rightarrow q$. Then we have the assertion $\lnot p$, and the conclusion therefore $ \lnot q$.

$$[(p \rightarrow q) \land \lnot p] \not \rightarrow \lnot q$$

Hence, this represents the fallacy of denying the hypothesis:

  • $p\rightarrow q$
  • $\lnot p$
  • $\therefore \lnot q$

which is not a valid inference.

The contrapositive of the first statement in (c) would be "If $n^2 \leq 4$, then $n \leq 2$, that is, if would be of the form $\lnot q \rightarrow \lnot p$.

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What two statements are you speaking about? I only see $[(P \implies q) \wedge \neg p]\implies \neg q$ – Mack Feb 8 '13 at 23:06
Let me know if this is any clearer now, Eli! – amWhy Feb 9 '13 at 2:33
+1 nice deduction. you did it well. – Babak S. Feb 9 '13 at 5:50
@amWhy Yes, it is, thank you. I invariably enjoy your answers. – Mack Feb 9 '13 at 12:31
@amWhy One more thing. I was wondering if this would be correct reasoning used, to arrive at the fact that part (c) is a fallacy. In the question, in the are asserting that if $p \rightarrow q$, and $\neg p$ is true, then these "true" statements assembled together imply that $\neg p \rightarrow \neg q$. However, this can't be so: a statement and its inverse can't both be true. Hence, it is a fallacy. – Mack Feb 9 '13 at 19:26

The argument is of the general shape

(1) If P, then Q.
(2) Not P.
(C) Therefore, not Q.

The important thing to learn is to that reasoning is not valid; indeed it fails to conclude truth from truth here in the particular case $n=-3$. Premise (1) is clearly a true statement about arithmetic; (2) is true here because $-3\le 2$, but the conclusion is false because $(-3)^2=9 > 4$.

I agree that it makes better sense to think of it as "using the contrapositive of (1) in the wrong direction" -- that is, the same basic fallacy as in (a), just with an intermediate correct step before -- than as a completely separate fallacy that needs a particular name of its own. It's not as if there's much practical benefit to reap from having such a fine-grained way to distinguish between different ways of being wrong.

I'd suggest not to spend too much time learning the fancy Latinate name for that particular fallacy, unless you need it to pass an exam that values rote learning of names over understanding; the important thing is just not to commit it. Very few mathematicians, even those who work in logic, bother to remember the details of the classical categorization of fallacies anyway.

It is somewhat more useful to remember the fancy names for valid reasoning, since they sometimes crop up in conversation about proof strategies and logical systems.

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In part (a), one tries to prove the converse which is not equivalent to the original.

In part (b), one tries to prove the contrapositive which is equivalent to the original.

In part (c), one tries to prove the inverse which is not equivalent to the original.

(For a conditional statement $a\rightarrow b$, its contrapositive is $\neg b\rightarrow\neg a$, its converse is $b\rightarrow a$, and its inverse is $\neg a\rightarrow\neg b$.)

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